# Question #1b9b0

Nov 22, 2016

Find $F ' \left(x\right)$

$F ' \left(x\right) = \frac{2 x \cos \left(2 x\right) - \sin \left(2 x\right)}{x} ^ 2$

Find when $F ' \left(x\right) = 0$ or undefined

$x \ne 0 \to$ Critical value
$x \approx 2.2467 , 3.8626 , 5.4521$$\in \left[0 , 2 \pi\right]$

Increasing on $\left(\approx 2.2467 , \approx 3.8626\right)$ & $\left(\approx 5.4521 , 2 \pi\right]$

hmmmmm

Nov 22, 2016

The graphing function for Socratic is currently not responding. I have done as much as I can without it. I leave the answer to you to work out the rest of, and will update this once the system is back in order.

#### Explanation:

$f \left(x\right) = \sin \frac{2 x}{x}$

Through the quotient rule,

$f ' \left(x\right) = \frac{2 x \cos \left(2 x\right) - \sin \left(2 x\right)}{x} ^ 2$

When $f ' > 0$, $f$ is increasing; when $f ' < 0$, $f$ is decreasing. We can not algebraically solve for when $f ' = 0$, so graph $f '$ to see when it is positive and negative.

$f '$ on $\left[0 , 2 \pi\right]$:

graph{(2xcos(2x)-sin(2x))/x^2 [-0.1, 6.3, -4.2, 5.8]}

$f$ is increasing when $f ' > 0$, which for the given interval is on $\left(2.247 , 3.863\right) \cup \left(5.452 , 2 \pi\right)$.

To find when $f$ is concave down, find when $f ' ' < 0$. Finding the second derivative is tedious, but when the quotient, product and chain rules are properly applied it should give:

$f ' ' \left(x\right) = - \frac{\left(4 {x}^{2} - 2\right) \sin \left(2 x\right) + 4 x \cos \left(2 x\right)}{x} ^ 3$

Graphing this to identify when it is negative, which corresponds to when $f$ is concave down:

$f ' '$ on $\left[0 , 2 \pi\right]$:

graph{-((4x^2-2)sin(2x)+4xcos(2x))/x^3 [-0.1, 6.3, -48.1, 49.8]}

Nov 22, 2016

See explanation and graph. $\pi = 3.1416$ in the graph.

#### Explanation:

graph{y-(sin (2x))/x=0 [-10, 10, -5, 5]}

sin 2x takes alternate signs in

$\left(0 , \frac{\pi}{2}\right) \left(\frac{\pi}{2} , \pi\right) \left(\pi , \frac{3}{2} \pi\right) \mathmr{and} \left(\frac{3}{2} \pi , 2 \pi\right)$,

The zeros of the function are at $x = \frac{\pi}{2} , \pi , \frac{3}{2} \pi \mathmr{and} 2 \pi$.

The concavity is up and down, alternately, at points of inflexion,

where the tangent crosses the curve. Seemingly, they are in

between zeros of f..

For precise locations, look for zeros of f''=o, by numerical iterative

methods. If I get these zeros, I would add them here in the next

edition. The graph easily reveals these tangent-crossing-curve

locations. But it lacks precision

he inserted Mountain-Hills range in the graph provides visual

effect. Nature is also a synonym to Mathematics that provide models

to natural phenomena.